﻿/*
描述
Given any string of N (>=5) characters, you are asked to form the characters into the shape of U. For example, "helloworld" can be printed as: h  d e  l l  r lowo That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 - 2 = N.
输入描述：
There are multiple test cases.Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.
输出描述：
For each test case, print the input string in the shape of U as specified in the description.
示例1
输入：
helloworld!
www.nowcoder.com
复制
输出：
h   !
e   d
l   l
lowor
w    m
w    o
w    c
.    .
n    r
owcode

*/



#include <iostream>
using namespace std;

int main()
{
	int n1, n2, n3;	//分别表示U的三个边的长度
	string str;
	char shape[28][28]; //(80 + 2) / 3 = 28;

	while (cin >> str)
	{
		int len = str.length();
		int n1 = n3 = (len + 2) / 3;	//减2从一开始就认为，其他两个边数值更小
		int n2 = len + 2 - n1 - n3;
		//cout << len + 2 << endl;
		//cout << n1 << endl << n3 << endl;
		//cout << n2 << endl;

		//在实际过程中，有时候需要对要输出的数组部分数据进行初始化，不然可能会因为数组里的乱码影响输出。比如'\0'和'\n'之类的
		for (int i = 0; i < n1; i++)
		{
			for (int j = 0; j < n2; j++)
			{
				shape[i][j] = ' ';
			}
		}
		
		for (int i = 0; i < n1-1; i++)
		{
			shape[i][0] = str[i];
			shape[i][n2 - 1] = str[len - i - 1];
		}

		for (int i = 0; i < n2; i++)
		{
			shape[n1 - 1][i] = str[i + n1 - 1];
		}

		for (int i = 0; i < n1; i++)
		{
			for (int j = 0; j < n2; j++)
			{
				cout << shape[i][j];
			}
			cout << endl;
		}

		//cout << str << endl;
	}
	return 0;
}

